International Commodity-Tax Competition and Asymmetric Producer Prices

Sezer Yasar

doi:10.1515/bejeap-2022-0332

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International Commodity-Tax Competition and Asymmetric Producer Prices

Sezer Yasar

Published/Copyright: January 13, 2025

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From the journal The B.E. Journal of Economic Analysis & Policy Volume 25 Issue 1

Abstract

In this paper, I extend the seminal commodity-tax competition model of Kanbur and Keen (1993. “Jeux Sans Frontières: Tax Competition and Tax Coordination When Countries Differ in Size.” The American Economic Review 83: 877–92) letting asymmetric producer prices between countries of different sizes. Unlike Kanbur and Keen, I show that there are multiple equilibria, and at some equilibria, small-country consumers cross-shop from the large. Besides, tax harmonization can benefit the small country, and a minimum tax rate can decrease a country’s tax revenue.

Keywords: commodity-tax competition; cross-border shopping; asymmetric prices; tax harmonization; minimum tax rate

JEL Classification: H20; H77

Corresponding author: Sezer Yasar, Cardiff Business School, Cardiff University, Cardiff, CF10 3EU, UK, E-mail: yasars3@cardiff.ac.uk

Acknowledgments

I thank the participants of TED University FEAS Brown Bag Seminar, 2019 International Conference of TED University Trade Research Center, and 2019 International Conference on Public Economic Theory for stimulating discussions. I also thank editor Tobias Wenzel and two anonymous referees for their valuable comments.

A Appendix: Proof of Lemmas and Propositions

A.1 Proof of Lemma 1

A.1.1 Small Country’s Best-Response Function

Assume that T ≤ p − P − δ. Then, whatever t is all small-country consumers shop from the large country. So, we have t(T) ∈ [0, ∞).

Assume that p − P − δ < T ≤ p − P. There is no tax rate such that the large-country consumers shop from the small country. Thus, the small country’s problem is max t t h 1 − ( t + p ) − ( T + P ) δ s.t. 0 ≤ t. Solving the problem, we have t ( T ) = δ + ( T + P ) − p 2 .

Assume that p − P < T. If the small country maximizes its revenue by choosing t ≤ (T + P) − p, its problem is given as max t t h + t H ( T + P ) − ( t + p ) δ s.t. 0 ≤ t ≤ (T + P) − p. Solving the problem, we have

(5) t = ( T + P ) − p if T < δ θ + p − P , δ θ + ( T + P ) − p 2 if δ θ + p − P ≤ T .

If the small country maximizes its revenue by choosing t ≥ T + P − p, its problem is given as max t t h 1 − ( t + p ) − ( T + P ) δ s.t. T + P − p ≤ t. Solving the problem, we have

(6) t = δ + ( T + P ) − p 2 if T ≤ δ + p − P , ( T + P ) − p if δ + p − P < T .

I analyze which way of maximizing its revenue is optimal for the small country under different values of θ. First, assume that θ < 1. Take T ∈ (p − P, δθ + p − P). If the small country maximizes its revenue by choosing t ≤ (T + P) − p, t = (T + P) − p by Equation (5), and its revenue is [(T + P) − p]h. If it maximizes its revenue by choosing t ≥ (T + P) − p, t = δ + ( T + P ) − p 2 by Equation (6), and its revenue is h ( δ + ( T + P ) − p ) 2 4 δ . We have h ( δ + ( T + P ) − p ) 2 4 δ > [ ( T + P ) − p ] h if and only if (δ − [(T + P) − p])² > 0 which always holds as we have δθ < δ and T < δθ + p − P. So, we have t ( T ) = δ + ( T + P ) − p 2 . Take T ∈ [δθ + p − P, δ + p − P]. If the small country maximizes its revenue by choosing t ≥ (T + P) − p, t = δ + ( T + P ) − p 2 by Equation (5), and its revenue is h ( δ + ( T + P ) − p ) 2 4 δ . If it maximizes its revenue by choosing t ≤ (T + P) − p, t = δ θ + ( T + P ) − p 2 by Equation (6), and its revenue is h ( δ θ + ( T + P ) − p ) 2 4 δ θ . We have h ( δ + ( T + P ) − p ) 2 4 δ > h ( δ θ + ( T + P ) − p ) 2 4 δ θ if and only if δ θ > ( T + P − p ) . So, we have

(7) t ( T ) = δ + ( T + P ) − p 2 if T < δ θ + p − P , δ + ( T + P ) − p 2 , δ θ + ( T + P ) − p 2 if T = δ θ + p − P .

Take T > δ + p − P. If the small country maximizes its revenue by choosing t ≤ (T + P) − p, t = δ θ + ( T + P ) − p 2 by Equation (5), and its revenue is h ( δ θ + ( T + P ) − p ) 2 4 δ θ . If it maximizes its revenue by choosing t ≥ (T + P) − p, t = (T + P) − p by Equation (6), and its revenue is [(T + P) − p]h. It is easy to show that the latter revenue is higher. Thus, t ( T ) = δ θ + ( T + P ) − p 2 . Collecting the results, if θ < 1, we have

(8) t ( T ) = [ 0 . ∞ ) if T ≤ p − P − δ δ + ( T + P ) − p 2 if p − P − δ < T < δ θ + p − P , δ + ( T + P ) − p 2 , δ θ + ( T + P ) − p 2 if T = δ θ + p − P , δ θ + ( T + P ) − p 2 if δ θ + p − P < T .

Second, assume that θ = 1. Take T ∈ (p − P, δ + p − P). If the small country maximizes its revenue by choosing t ≤ (T + P) − p, t = (T + P) − p by Equation (5), and its revenue is [(T + P) − p]h. If it maximizes its revenue by choosing t ≥ (T + P) − p, t = δ + ( T + P ) − p 2 by Equation (6), and its revenue is h ( δ + ( T + P ) − p ) 2 4 δ . We have h ( δ + ( T + P ) − p ) 2 4 δ > [ ( T + P ) − p ] h if and only if (δ − [(T + P) − p])² > 0 which always holds as we have θ = 1 and T < δθ + p − P. So, we have t ( T ) = δ + ( T + P ) − p 2 . Take T = δθ + p − P = δ + p − P. If the small country maximizes its revenue by choosing t ≤ (T + P) − p, t = δ θ + ( T + P ) − p 2 by Equation 5, and if it maximizes its revenue by choosing t ≥ (T + P) − p, t = δ + ( T + P ) − p 2 by Equation (6). Because, both tax rates are equal, as θ = 1, we have t ( T ) = δ + ( T + P ) − p 2 . Take T > δ + p − P. If the small country maximizes its revenue by choosing t ≤ (T + P) − p, t = δ θ + ( T + P ) − p 2 by Equation (5), and its revenue is h ( δ θ + ( T + P ) − p ) 2 4 δ θ . If it maximizes its revenue by choosing t ≥ (T + P) − p, t = (T + P) − p by Equation (6), and its revenue is [(T + P) − p]h. It is easy to show that the former revenue is higher. Thus, we have t ( T ) = δ + ( T + P ) − p 2 . Collecting the results, if θ = 1, we have

(9) t ( T ) = [ 0 . ∞ ) if T ≤ p − P − δ δ + ( T + P ) − p 2 otherwise .

The whole results give us the small country’s best-response function:

t ( T ) = [ 0 , ∞ ) if T ≤ p − P − δ , δ + P + T − p 2 if p − P − δ < T < δ θ + p − P , δ + P + T − p 2 , δ θ + P + T − p 2 if T = δ θ + p − P , δ θ + P + T − p 2 if δ θ + p − P < T .

A.1.2 Large Country’s Best-Response Function.

Assume that t ≤ −(p − P) − δ. Then, whatever T is all large-country consumers shop from the small country. So, we have T(t) ∈ [0, ∞).

Assume that −(p − P) − δ < t ≤ −(p − P). There is no tax rate such that the small-country consumers shop from the large country. Thus, the large country’s problem is max T T H 1 − ( T + P ) − ( t + p ) δ s.t. 0 ≤ T. Solving the problem, we have T ( t ) = δ + ( t + p ) − P 2 .

Assume that −(p − P) < t. If the large country maximizes its revenue by choosing T ≤ (t + p) − P, its problem is given as max T T H + T h ( t + p ) − ( T + P ) δ s.t. 0 ≤ T ≤ (t + p) − P. Solving the problem, we have

(10) T = ( t + p ) − P if t < δ 1 θ − ( p − P ) , δ 1 θ + ( t + p ) − P 2 if δ 1 θ − ( p − P ) ≤ t .

If the large country maximizes its revenue by choosing T ≥ t + p − P, its problem is given as max T T H 1 − ( T + P ) − ( t + p ) δ s.t. t + p − P ≤ T. Solving the problem, we have

(11) T = δ + ( t + p ) − P 2 if t ≤ δ − ( p − P ) , ( t + p ) − P if δ − ( p − P ) < t .

I analyze which way of maximizing its revenue is optimal for the small country under different values of θ.

First, assume that θ = 1. Then, clearly, the best-response function of the large country is analogous to the small country. So, we have

(12) T ( t ) = [ 0 . ∞ ) if t ≤ − ( p − P ) − δ δ + ( t + p ) − P 2 otherwise .

Second, assume that 1 < θ. Take t ∈ (−(p − P), δ − (p − P)). If the large country maximizes its revenue by choosing T ≤ (t + p) − P, T = (t + p) − P by Equation (10), and its revenue is [(t + p) − P]H. If it maximizes its revenue by choosing T ≥ (t + p) − P, T = δ + ( t + p ) − P 2 by Equation (11), and its revenue is H ( δ + ( t + p ) − P ) 2 4 δ . It is easy to show that the former revenue is higher. Thus, we have T ( t ) = δ + ( t + p ) − P 2 . Take t ∈ [ δ − ( p − P ) , δ 1 θ − ( p − P ) ] . Whether the large country maximizes its revenue by choosing T ≤ (t + p) − P or T ≥ (t + p) − P, it is optimal to choose T = (t + p) − P by Equations (10) and (11). So, we have T(t) = (t + p) − P. Take t > δ 1 θ − ( p − P ) . If the large country maximizes its revenue by choosing T ≤ (t + p) − P, T = δ 1 θ + ( t + p ) − P 2 by Equation (10), and its revenue is H ( δ 1 θ + ( t + p ) − P ) 2 4 δ 1 θ . If it maximizes its revenue by choosing T ≥ (t + p) − P, T = (t + p) − P by Equation (11), and its revenue is [(t + p) − P]H. It is easy to show that the latter revenue is higher if and only if t ≤ δ 1 θ − ( p − P ) . Collecting the results, we have

(13) T ( t ) = [ 0 , ∞ ) if t ≤ − ( p − P ) − δ , δ + p + t − P 2 if − ( p − P ) − δ < t ≤ δ − ( p − P ) , p + t − P if δ − ( p − P ) < t ≤ δ 1 θ − ( p − P ) , δ 1 θ + p + t − P 2 if δ 1 θ − ( p − P ) < t .

□

A.2 Proof of Proposition 1

In the following, I assume that θ ≤ 1. So, we have

t ( T ) = [ 0 , ∞ ) if T ≤ p − P − δ , δ + P + T − p 2 if p − P − δ < T < δ θ + p − P , δ + P + T − p 2 , δ θ + P + T − p 2 if T = δ θ + p − P , δ θ + P + T − p 2 if δ θ + p − P < T ,

and

(14) T ( t ) = [ 0 , ∞ ) if t ≤ − ( p − P ) − δ , δ + ( t + p ) − P 2 if − ( p − P ) − δ < t ≤ δ − ( p − P ) , ( t + p ) − P if δ − ( p − P ) < t ≤ δ 1 θ − ( p − P ) , δ 1 θ + ( t + p ) − P 2 if δ 1 θ − ( p − P ) < t .

To prove the proposition, I present both countries’ best-response correspondences and find their intersection points for each part of the proposition. Additionally, Part (iii) of the proposition is implied by Parts (ii) and (iv). Thus, I give the proof of Part (iii) after the proof of Part (iv).

Part i: Assume that (p − P) ≤ −2δ − δθ. Then, the best-response correspondences of the small and large countries are as given in Figure 7. Under the parameter restrictions of Part (i), it is easy to show that the best-response correspondences intersect as in Figure 8. Solving for the intersection points, we get a continuum of equilibria such that T N ∈ 0 , − 2 δ − δ θ − ( p − P ) and t N = δ θ + T N − ( p − P ) 2 .

Figure 7:

Best-response correspondences in the proof of Proposition 1 Part (i) and (ii) when −2δ − δθ < p − P ≤ −δ.

Figure 8:

Equilibria in the proof of Proposition 1 Part (i).

Part ii: I prove Part (ii) of the proposition under three cases, as the shape of the best-response correspondences differs. First, assume that −2δ − δθ < p − P ≤ −δ. Then, the best-response correspondences of the small and large countries are given in Figure 7. Under the parameter of our restrictions, it is easy to show that the best-response correspondences intersect as in Figure 9. Solving for the intersection points, we get the unique equilibrium ( t N , T N ) = 2 δ θ + δ − ( p − P ) 3 , 2 δ + δ θ + ( p − P ) 3 .

Figure 9:

Equilibrium in the proof of Proposition 1 Part (ii) when −2δ − δθ < p − P ≤ −δ.

Second, assume that − δ < p − P ≤ − δ θ . Then, the best-response correspondences of the small and large countries are given in Figure 10. It is easy to show that the best-response correspondences intersect only if t ≤ δ − (p − P). For the other values of t, if solve for an intersection point, given our assumption on the parameters, the values of t or T fall out of the interval that we assume for it. If we solve for an intersection point when t ≤ δ − (p − P), we get the unique point ( t N , T N ) = 2 δ θ + δ − ( p − P ) 3 , 2 δ + δ θ + ( p − P ) 3 . We can show that, given our assumption on the parameters, t _N ≤ δ − (p − P) and T _N ≥ 0. Thus, this intersection point is the unique equilibrium.

$Figure 10: Best-response correspondences in the proof of Proposition 1 Part (ii) when − δ < p − P ≤ − δ θ $-\delta {< }p-P\le -\delta \sqrt{\theta }$ .$

Figure 10:

Best-response correspondences in the proof of Proposition 1 Part (ii) when − δ < p − P ≤ − δ θ .

Third, assume that − δ θ . < ( p − P ) ≤ δ . Then, the best-response correspondences of the small and large countries are given in Figure 11. It is easy to show that the best-response correspondences intersect only if t ≤ δ − (p − P) and δ θ + p − P ≤ T . In other cases, if we solve for an intersection point, given our assumption on the parameters, the values of t or T fall out of the interval that we assume for it. If we solve for an intersection point when t ≤ δ − (p − P) and δ θ + p − P ≤ T , we get the unique point ( t N , T N ) = 2 δ θ + δ − ( p − P ) 3 , 2 δ + δ θ + ( p − P ) 3 . We need t _N ≤ δ − (p − P) and δ θ + p − P ≤ T N which require p − P ≤ 2 δ + δ θ − 3 δ θ 2 .

$Figure 11: Best-response correspondences in the proof of Proposition 1 Part (ii) when − δ θ ≤ ( p − P ) ≤ δ $-\delta \sqrt{\theta }\le \left(p-P\right)\le \delta $ .$

Figure 11:

Best-response correspondences in the proof of Proposition 1 Part (ii) when − δ θ ≤ ( p − P ) ≤ δ .

So, collecting the results, if − δ < p − P ≤ 2 δ + δ θ − 3 δ θ 2 , we have a unique equilibrium ( t N , T N ) 2 δ θ + δ − ( p − P ) 3 , 2 δ + δ θ + ( p − P ) 3 .

Part iii. Proof Part (iii) is given after proof of Part (iv) as it is implied by the proofs of Part (ii) and (iv).

Part iv: I prove Part (iv) of the proposition under two cases, as the shape of the best-response correspondences differs under the two. First, assume that δ ≤ p − P < δ 1 θ . Then, the best-response correspondences of the small and large countries are given in Figure 12. It is easy to show that the best-response correspondences intersect only if δ 1 θ − ( p − P ) ≤ t and p − P − δ ≤ T ≤ δ θ + p − P . In other cases, if solve for an intersection point, given our assumption on the parameters, the values of t or T fall out of the interval that we assume for it. If we solve for an intersection point when δ 1 θ − ( p − P ) ≤ t and p − P − δ ≤ T ≤ δ θ + p − P , we get the unique point ( t N , T N ) = 2 δ + δ 1 θ − ( p − P ) 3 , 2 δ 1 θ + δ + p − P 3 . We need δ 1 θ − ( p − P ) ≤ t N and p − P − δ ≤ T N ≤ δ θ + p − P which require 2 δ 1 θ + δ − 3 δ θ 2 ≤ p − P .

$Figure 12: Best-response correspondences in the proof of Proposition 1 Part (iv) when δ < p − P < δ 1 θ $\delta {< }p-P{< }\delta \frac{1}{\theta }$ .$

Figure 12:

Best-response correspondences in the proof of Proposition 1 Part (iv) when δ < p − P < δ 1 θ .

Second, assume that δ 1 θ ≤ ( p − P ) ≤ δ 1 θ + 2 δ . Then, the best-response correspondences of the small and large countries are given in Figure 13. It is easy to show that the best-response correspondences intersect only if p − P − δ ≤ T ≤ δ θ + p − P . For the other ranges of T, if solve for an intersection point, given our assumption on the parameters, the values of t or T fall out of the interval that we assume for it. If we solve for an intersection point when p − P − δ ≤ T ≤ δ θ + p − P , we get the unique point ( t N , T N ) = 2 δ + δ 1 θ − ( p − P ) 3 , 2 δ 1 θ + δ + p − P 3 . We need p − P − δ ≤ T N ≤ δ θ + p − P which requires 2 δ 1 θ + δ − 3 δ θ 2 ≤ p − P .

$Figure 13: Best-response correspondences in the proof of Proposition 1 Part (iv), when δ 1 θ ≤ ( p − P ) ≤ δ 1 θ + 2 δ $\delta \frac{1}{\theta }\le \left(p-P\right)\le \delta \frac{1}{\theta }+2\delta $ , and in Part (v).$

Figure 13:

Best-response correspondences in the proof of Proposition 1 Part (iv), when δ 1 θ ≤ ( p − P ) ≤ δ 1 θ + 2 δ , and in Part (v).

Collecting the results in both parts of the proof, if max δ , 2 δ 1 θ + δ − 3 δ θ 2 ≤ p − P ≤ δ 1 θ + 2 δ , we have a unique equilibrium ( t N , T N ) = 2 δ + δ 1 θ − ( p − P ) 3 , 2 δ 1 θ + δ + p − P 3 .

Part iii. By the second part of the proof of Part (ii), there is no equilibrium if 2 δ + δ θ − 3 δ θ 2 < p − P < δ . By the proof of Part (iv), there is no equilibrium if δ < p − P < max δ , 2 δ 1 θ + δ − 3 δ θ 2 . Combining the two results, there is no equilibrium if 2 δ + δ θ − 3 δ θ 2 < p − P < max δ , 2 δ 1 θ + δ − 3 δ θ 2 .

Part v: Assume that δ 1 θ + 2 δ < p − P . Then, the best-response correspondences of the small and large countries are as given in Figure 13. Then, under the assumptions of Part (i), it is easy to show that the best-response correspondences intersect as in Figure 14. Solving for the intersection points, we get a continuum of equilibria such that t N ∈ 0 , p − P − 2 δ − δ 1 θ and T N = δ 1 θ + p − P + t N 2 . □

Figure 14:

Equilibria in the proof of Proposition 1 Part (v).

A.3 Proof of Proposition 2

We have cross-border sales from small to the large country if and only if t _N + p < T _N + P. Rewriting the condition under Type A equilibrium, by Proposition 1 Part (ii), we get p − P < δ(1 − θ). By the same proposition, we have Type A equilibrium if and only if − δ < p − P ≤ 2 δ + δ θ − 3 δ θ 2 and 2 δ + δ θ − 3 δ θ 2 < δ ( 1 − θ ) , the condition for cross-border sales from small to large country always holds.

By Proposition 1 Part (ii), we have t _N < T _N under Type A equilibrium if and only if 2 δ θ + δ − ( p − P ) 3 < 2 δ + δ θ + ( p − P ) 3 , that is − 1 2 δ ( 1 − θ ) < p − P . □

A.4 Proof of Proposition 3

We have cross-border sales from the large country to the small country if and only if T _N + P < t _N + p. Rewriting the condition using Proposition 1 Part (iv), we have δ 1 θ − 1 < p − P . We have Type B equilibrium if and only if max δ , 2 δ 1 θ + δ − 3 δ θ 2 ≤ p − P ≤ δ 1 θ + 2 δ . If max δ , 2 δ 1 θ + δ − 3 δ θ 2 = δ , we have δ 1 θ − 1 < δ ≤ p − P and so the condition for cross-border sales from the large country is satisfied. If max δ , 2 δ 1 θ + δ − 3 δ θ 2 = 2 δ 1 θ + δ − 3 δ θ 2 , we have δ 1 θ − 1 < 2 δ 1 θ + δ − 3 δ θ 2 ≤ p − P and so the condition for cross-border sales from the large country is satisfied. Thus, there are always cross-border sales from the large to the small country under a Type B equilibrium.

By Proposition 1 Part (iv), we have t _N < T _N if and only if − 1 2 δ 1 θ − 1 < p − P which always holds as under a Type B equilibrium we have 0 ≤ p − P by Proposition 1 Part (iv). □

A.5 Proof of Proposition 4

I prove the proposition under four cases depending on the parameter values.

Case 1: Assume that − 2 δ − δ θ < p − P < − 1 2 δ ( 1 − θ ) . We have cross-border sales in the small country and T _N < t _N by Proposition 2. Additionally, the tax revenues are given as R ( t N , T N ) = T N H 1 − ( P + T N ) − ( p + t N ) δ and r ( t N , T N ) = t N h + t N H ( P + T N ) − ( p + t N ) δ by Proposition 1 Part (ii) and Proposition 2. Under tax harmonization, take τ ∈ [T _N, t _N]. Then, it is easy to show that we have cross-border sales in the small country, and the tax revenues are given as R ( τ ) = τ H 1 − P − p δ and r ( τ ) = τ h + τ H P − p δ .

Part i: We have R(t _N) = R(t _N, t _N) < R(t _N, T _N) as ( P + T N ) − ( p + t N ) δ < P − p δ . Thus, considering that R(τ) is increasing, we have R(τ) < R(t _N, T _N) for all τ ∈ [T _N, t _N].

Part ii: We have r(t _N, T _N) < r(t _N) as ( P + T N ) − ( p + t N ) δ < P − p δ . We also have r(T _N) = r(T _N, T _N) < r(t _N, T _N). Thus, considering that r(τ) is increasing, there exits τ ̄ such that r(t _N, T _N) < r(τ) if and only if τ ̄ < τ .

Case 2: Assume that − 1 2 δ ( 1 − θ ) < p − P < 0 . We have cross-border sales in the small country and t _N < T _N by Proposition 2. Additionally, the tax revenues are given as R ( t N , T N ) = T N H 1 − ( P + T N ) − ( p + t N ) δ and r ( t N , T N ) = t N h + t N H ( P + T N ) − ( p + t N ) δ by Proposition 1 Part (ii) and Proposition 2. Under tax harmonization, take τ ∈ [t _N, T _N]. Then, it is easy to show that we have cross-border sales in small the country, and the tax revenues are given as R ( τ ) = τ H 1 − P − p δ and r ( τ ) = τ h + τ H P − p δ .

Part i: We have r(T _N) = r(T _N, T _N) < r(t _N, T _N) as P − p δ < ( P + T N ) − ( p + t N ) δ . We also have r(T _N) = r(T _N, T _N) < r(t _N, T _N). Thus, considering that r is increasing, r(τ) < r(t _N, T _N) for all τ ∈ [t _N, T _N].

Part ii: We have R(T _N) > R(t _N, T _N) as P − p δ < ( P + T N ) − ( p + t N ) δ . We also have R(t _N) = R(t _N, t _N) < R(t _N, T _N). Thus, considering that R is increasing, there exists τ ̄ ∈ [ t N , T N ] such that R(t _N, T _N) < R(τ) if and only if τ ̄ < τ .

Case 3: Assume that 0 < p − P ≤ 2 δ + δ θ − 3 δ θ 2 . We have cross-border sales in the small country and t _N < T _N by Proposition 2. Additionally, the tax revenues are given as R ( t N , T N ) = T N H 1 − ( P + T N ) − ( p + t N ) δ and r ( t N , T N ) = t N h + t N H ( P + T N ) − ( p + t N ) δ by Proposition 1 Part (ii) and Proposition 2. Under tax harmonization, take τ ∈ [t _N, T _N]. Then, it is easy to show that we have cross-border sales in the large country, and the tax revenues are given as R ( τ ) = τ H + τ h p − P δ and r ( τ ) = τ h 1 − p − P δ .

Part i: We have r(T _N) = r(T _N, T _N) < r(t _N, T _N). Thus, considering that r is increasing, r(τ) < r(t _N, T _N) for all τ ∈ [t _N, T _N].

Part ii: We have R(T _N) > R(t _N, T _N) as p − P δ > p + t N − ( P + T N ) δ . We also have R(t _N) = R(t _N, t _N) < R(t _N, T _N). Thus, considering that R is increasing, there exists τ ̄ ∈ [ t N , T N ] such that R(t _N, T _N) < R(τ) if and only if τ ̄ < τ .

Case 4: Assume that max δ , 2 δ 1 θ + δ − 3 δ θ 2 ≤ p − P ≤ δ 1 θ + 2 δ . We have cross-border sales in the large country and t _N < T _N by Proposition 3. We have r(T _N) = r(T _N, T _N) < r(t _N, T _N). Additionally, the tax revenues are given as R ( t N , T N ) = T N H + T N h p + t N − ( P + T N ) δ and r ( t N , T N ) = t N h 1 − p + t N − ( P + T N ) δ by Proposition 1 Part (iv) and Proposition 3. Under tax harmonization, take τ ∈ [t _N, T _N]. Then, it is easy to show that we have cross-border sales in the large country, and R ( τ ) = τ H + τ h p − P δ and r ( τ ) = τ h 1 − p − P δ .

Part i: We have r(T _N) = r(T _N, T _N) < r(t _N, T _N). Thus, considering that r is increasing, r(τ) < r(t _N, T _N) for all τ ∈ [t _N, T _N].

A.6 Proof of Proposition 5

Under a minimum-tax rate, I show the equilibrium tax rates by t _m and T _m for the small and large countries, respectively.

Part i: Assume that − δ θ < p − P < 2 δ + δ θ − 3 δ θ 2 . Then, we have t N = 2 δ θ + δ − ( p − P ) 3 < 2 δ + δ θ + ( p − P ) 3 = T N by Proposition 1 Part (ii), and cross-shopping in the small country by Proposition 2. These imply that we have R ( t N , T N ) = H 1 δ δ θ + 2 δ + p − P 3 2 and r ( t N , T N ) = H 1 δ 2 δ θ + δ − ( p − P ) 3 2 . Assume further that 1 4 δ ( 1 − θ ) < p − P < 2 δ + δ θ − 3 δ θ 2 . so that we have δ − ( p − P ) < T N < δ 1 θ − ( p − P ) . Then, the equilibrium is as given in Figure 15.

Figure 15:

Equilibrium in the proof of Proposition 5 Part (i).

Take the minimum tax rate μ ∈ δ − ( p − P ) , T N . Then, we have t _m = μ < μ + p − P = T _m. There are cross-border sales if and only if μ + t _m ≠ T _m + P, that is, μ ≠ μ which never holds. So, there is no cross-border sale under the minimum tax rate.

We have R(μ) = (μ + p − P)H. Then R(μ) > R(t _N, T _N) if and only if ( μ + p − P ) H > T H 1 − T + P − ( t + p ) δ , that is, δ ( μ + p − P ) > δ θ + 2 δ + p − P 3 2 . For the minimum value of μ, the left-hand side of this inequality is δ ². Additionally, we have δ 2 > δ θ + 2 δ + p − P 3 2 if and only if p − P < δ(1 − θ) which always holds as 2 δ + δ θ − 3 δ θ 3 < δ ( 1 − θ ) .

We have r(μ) = μh. Then, r(μ) > r(t _N, T _N) if and only if μ h > H 1 δ 2 δ θ + δ − ( p − P ) 3 2 , that is, μ δ θ > 2 δ θ + δ − ( p − P ) 3 2 . It is easy to show that the graph of both sides of this inequality is as given in Figure 16. So, there exists (p − P)_m and (p − P)_M such that a minimum tax rate increases the small country’s tax revenue if and only if (p − P)_m < p − P < (p − P)_M.

Figure 16:

Graph of the inequality in the proof of Proposition 5 Part (i).

Part ii: I prove Part (ii) of the proposition by considering different cases for the value of p − P, as the figure of the best-response correspondences and the range of the feasible values of the minimum-tax rate differ under each case.

Case 1: Assume that −2δ − δθ < p − P ≤ δ. Then, under a minimum tax rate μ ∈ [T _n, t _n], the best-response correspondences are given in Figure 17. As shown by the dashed lines and the lightly shadowed area, both countries’ best-response correspondences are irrelevant for the other country’s tax rate below μ. Then, as seen in the graph, the best-response correspondences do not intersect and we do not have an equilibrium. So, I continue to the proof, by focusing on the remaining cases.

Figure 17:

Best-response correspondences in Case 1 of proof of Proposition 5 Part (ii).

Case 2: Assume that − δ < p − P < min − δ θ , − 1 2 δ ( 1 − θ ) . Then, we have T N = 2 δ + δ θ + ( p − P ) 3 < 2 δ θ + δ − ( p − P ) 3 = t N by Proposition 1 Part (ii), and cross-shopping in the small country by Proposition 2. These imply that we have R ( t N , T N ) = H 1 δ δ θ + 2 δ + p − P 3 2 and r ( t N , T N ) = H 1 δ 2 δ θ + δ − ( p − P ) 3 2 and the equilibrium is as given in Figure 18.

Figure 18:

Equilibrium in Case 2 of proof of Proposition 5 Part (ii).

Following the discussion in Section 4, under a minimum-tax rate μ ∈ [T _N, t _N], we can ignore T(t) for t < μ. Additionally, as T can not be less than μ, for values of t > μ, we have T = μ until the value of t is such that T(t) = μ. And t(T) will get longer before the discontinuity point and will be equal to μ before the value of T is such that t(T) = μ. Then, we have T m = μ < δ θ + μ − ( p − P ) 2 = t m . Under the minimum tax rate, there are cross-border sales in the small country if and only if t _m + p < T _m + P, that is, δθ + (p − P) < μ which always holds as μ ≥ T _N.

We have R ( μ ) = T m H 1 − T m + P − ( t m + p ) δ = μ H 2 δ − μ + ( p − P ) + δ θ 2 δ . Notice that R(μ) > R(t _N, T _N) if and only if δ θ + 2 δ + ( p − P ) μ − μ 2 2 > δ θ + 2 δ + p − P 2 9 . It is easy to show that the graph of the left-hand side of this inequality is as given in Figure 19. Then, the minimum tax rate increases the large country’s tax revenue if and only if δ θ + 2 δ + ( p − P ) 3 < μ < δ θ + 2 δ + ( p − P ) 2 + δ θ + 2 δ + ( p − P ) 2 − δ θ + 2 δ + ( p − P ) 3 , that is, δ θ + 2 δ + ( p − P ) 3 < μ < 2 [ δ θ + 2 δ + ( p − P ) ] 3 which always holds as T N < μ < t N < 2 [ δ θ + 2 δ + ( p − P ) ] 3 .

Figure 19:

Graph of the inequality in Case 1 of proof of Proposition 5 Part (ii).

We have r ( μ ) = t m h + t m H T m + P − ( t m + p ) δ = 1 δ H δ θ + μ − ( p − P ) 2 2 . Then, r(μ) > r(t _N, T _N) if and only if μ > δ θ + 2 δ + ( p − P ) 3 which always holds as μ > T _N.

So, for Case 2, we have μ ̄ = t N .

Case 3: Assume that − 1 2 δ ( 1 − θ ) < p − P < − δ θ . Then, we have t N = 2 δ θ + δ − ( p − P ) 3 < 2 δ + δ θ + ( p − P ) 3 = T N by Proposition 1 Part (ii), and cross-shopping in the small country by Proposition 2. These imply that we have R ( t N , T N ) = H 1 δ δ θ + 2 δ + p − P 3 2 and r ( t N , T N ) = H 1 δ 2 δ θ + δ − ( p − P ) 3 2 and the equilibrium is as given in Figure 20.

Figure 20:

Equilibrium Case 3 of proof of Proposition 5 Part (ii).

Following the discussion in Section 4, under a minimum-tax rate μ ∈ [t _N, T _N], we can ignore T(t) for t < μ. And t(T) will get longer before the discontinuity point and will be equal to μ before the value of T such that t(T) = μ. Then, we have t m = μ < δ + μ + ( p − P ) 2 = T m . There are cross-border sales in the small country if and only if t _m + p < T _m + P, that is, p − P < δ − μ. Notice that we have δ − T _N < δ − μ < δ − t _N, that is, δ − 2 δ + δ θ + ( p − P ) 3 < δ − μ < δ − 2 δ θ + δ + ( p − P ) 3 . By the left-hand side of this inequality, we have δ − δ θ − ( p − P ) 3 < δ − μ . Assume that we have δ − δ θ − ( p − P ) 3 < p − P . This implies that 1 4 δ ( 1 − θ ) < p − P which contradicts our assumption for Case 3. So, there are always cross-border sales in the small country.

We have R ( μ ) = T m H 1 − T m + P − ( t m + p ) δ = H 1 δ μ + δ + ( p − P ) 2 2 . Notice that R(μ) > R(t _N, T _N) if and only if H 1 δ μ + δ + ( p − P ) 2 2 > H 1 δ δ θ + 2 δ + p − P 3 2 , that is, μ > 2 δ θ + δ − ( p − P ) 3 = t N which always holds as μ > t _N.

We have r ( μ ) = t m h + t m H T m + P − ( t m + p ) δ = H 1 δ − μ 2 + 2 δ θ + δ − ( p − P μ 2 . Then, r(μ) > r(t _N, T _N) if and only if − μ 2 + 2 δ θ + δ − ( p − P μ 2 > 2 δ θ + δ − ( p − P ) 3 2 . It is easy to show that the graph of the left-hand side of the above inequality is given in Figure 21. Then, the minimum tax rate increases the small country’s tax revenue if and only if 2 δ θ + δ − ( p − P ) 3 < μ < 2 δ θ + δ − ( p − P ) 2 + 2 δ θ + δ − ( p − P ) 2 − 2 δ θ + δ − ( p − P ) 3 that is 2 δ θ + δ − ( p − P ) 3 < μ < 2 [ 2 δ θ + δ − ( p − P ) ] 3 which always holds as t N < μ < T N < 2 [ 2 δ θ + δ − ( p − P ) ] 3 .

Figure 21:

Graph of the inequality in Case 3 of proof of Proposition 5 Part (ii).

So, for Case 3, we have μ ̄ = T N .

Case 4: Assume that − δ θ < p − P < 2 δ + δ θ − 3 δ θ 2 . Then, we have t N = 2 δ θ + δ − ( p − P ) 3 < 2 δ + δ θ + ( p − P ) 3 = T N by Proposition 1 Part (ii), and cross-shopping in the small country by Proposition 2. These imply that we have R ( t N , T N ) = H 1 δ δ θ + 2 δ + p − P 3 2 and r ( t N , T N ) = H 1 δ 2 δ θ + δ − ( p − P ) 3 2 , and the equilibrium given in Figure 22.

Figure 22:

Equilibrium in Case 3 of proof of Proposition 5 Part (ii).

Case 4.i: Assume that − δ θ < p − P < 1 4 δ ( 1 − θ ) so that we have T _N < δ − (p − P). Take the minimum tax rate μ ∈ t N , T N . Then, we have t m = μ < δ + μ + ( p − P ) 2 = T m . We have cross-border sales in the small country if and only if t _m + p < T _m + P, that is, μ < δ − (p − P) which always holds as μ < T _N < δ − (p − P). So, there are always cross-border sales in the small country.

We have r ( μ ) = t m h + t m H T m + P − ( t m + p ) δ = H 1 δ − μ 2 + 2 δ θ + δ − ( p − P μ 2 . Then r(μ) > r(t _N, T _N) if and only if H 1 δ − μ 2 + 2 δ θ + δ − ( p − P μ 2 > H 1 δ 2 δ θ + δ − ( p − P ) 3 2 , that is, − μ 2 + 2 δ θ + δ − ( p − P μ 2 > 2 δ θ + δ − ( p − P ) 3 2 . It is easy to show that the graph of the left-hand side of this inequality is given in Figure 23. Then, the minimum tax rate increases the small country’s tax revenue if and only if 2 δ θ + δ − ( p − P ) 3 < μ < 2 δ θ + δ − ( p − P ) 2 + 2 δ θ + δ − ( p − P ) 2 − 2 δ θ + δ − ( p − P ) 3 , that is, 2 δ θ + δ − ( p − P ) 3 < μ < 2 [ 2 δ θ + δ − ( p − P ) ] 3 .

Figure 23:

Graph of the inequality in Case 4(i) of the proof of Proposition 5 Part (ii).

So, for Case 4.i, we have μ ̄ = 2 [ 2 δ θ + δ − ( p − P ) ] 3 .

Case 4.ii: Assume that 1 4 δ ( 1 − θ ) < p − P < 2 δ + δ θ − 3 δ θ 2 . so that we have δ − ( p − P ) < T N < δ 1 θ − ( p − P ) . Take the minimum tax rate μ ∈ t N , δ − ( p − P ) . Then, we have t m = μ < δ + μ + ( p − P ) 2 = T m . There is still a cross-border sale in the small country if and only if t _m + p < T _m + P, that is, p − P < δ − μ. Notice that by our assumption about the parameters, in this case, we have δ − ( δ − ( p − P ) ) < δ − μ < δ − 2 δ θ + δ + ( p − P ) 3 , that is, p − P < δ − μ < 2 δ − 2 δ θ + p − P 3 . So, we always have a cross-border sale in the small country.

We have r ( μ ) = t m h + t m H T m + P − ( t m + p ) δ = H 1 δ − μ 2 + 2 δ θ + δ − ( p − P μ 2 . Then, r(μ) > r(t _N, T _N) if and only if H 1 δ − μ 2 + 2 δ θ + δ − ( p − P μ 2 > H 1 δ 2 δ θ + δ − ( p − P ) 3 2 , that is, − μ 2 + 2 δ θ + δ − ( p − P μ 2 > 2 δ θ + δ − ( p − P ) 3 2 . It is easy to show that the graph of the left-hand side of this inequality is given in Figure 24. Then, the minimum tax rate increases the small country’s tax revenue if and only if 2 δ θ + δ − ( p − P ) 3 < μ < 2 δ θ + δ − ( p − P ) 2 + 2 δ θ + δ − ( p − P ) 2 − 2 δ θ + δ − ( p − P ) 3 , that is, 2 δ θ + δ − ( p − P ) 3 < μ < 2 [ 2 δ θ + δ − ( p − P ) ] 3 .

Figure 24:

Graph of the inequality in Case 4(ii) of proof of Proposition 5 Part (ii).

So, for Case 4.ii, we have μ ̄ = 2 [ 2 δ θ + δ − ( p − P ) ] 3 .

Case 5: Assume that max δ , 2 δ 1 θ + δ − 3 δ θ 2 < p − P < δ 1 θ . Then, we have t N = 2 δ + δ 1 θ − ( p − P ) 3 < T N = 2 δ 1 θ + δ + p − P 3 by Proposition 1 Part (iv), and cross-shopping in the large country by Proposition 2. These imply that we have R ( t N , T N ) = h 1 δ 2 δ 1 θ + δ + p − P 3 2 and r ( t N , T N ) = h 1 δ 2 δ + δ 1 θ − ( p − P ) 3 2 and the equilibrium given in Figure 25. Following the discussion in Section 4, under a minimum-tax rate μ ∈ [t _N, T _N], we can ignore T(t) for t < μ. And t(T) will be equal to μ before the μ-line cuts t(T) in its middle part, get longer before the discontinuity point, and will be equal to μ before the value of T is such that t(T) = μ. Then, we have t m = μ < δ 1 θ + μ + ( p − P ) 2 = T m . There is a cross-border sale in the large country if and only if T _m + P < t _m + p, that is, δ 1 θ − ( p − P ) < μ . Assume that μ < δ 1 θ − ( p − P ) so that there is a cross-border sale in the small country. For this to hold, we need t N < δ 1 θ − ( p − P ) which implies that p − P < δ 1 θ − δ . For this to hold, we need 2 δ 1 θ + δ − 3 δ θ 2 < δ 1 θ − δ , that is, 1 < θ which never holds. So, there is always cross-border sale in the large country.

Figure 25:

Equilibrium in Case 5 of proof of Proposition 5 Part (ii).

We have R ( μ ) = T m H + T m h t m + p − ( T m + P ) δ = h 1 δ δ 1 θ + μ + ( p − P ) 2 2 . Then, R(μ) ≥ R(t _N, T _N) if and only if h 1 δ δ 1 θ + μ + ( p − P ) 2 2 ≥ h 1 δ 2 δ 1 θ + δ + p − P 3 2 , that is, μ ≥ δ 1 θ + 2 δ − ( p − P ) 3 = T N which always holds as μ > T _N.

We have r ( μ ) = t m h 1 − t m + p − ( T m + P ) δ = h 1 δ μ 2 δ − μ + δ 1 θ − ( p − P ) 2 . Then, r ( μ ) ≥ r ( t N , T ) N if and only if h 1 δ μ 2 δ − μ + δ 1 θ − ( p − P ) 2 > h 1 δ 2 δ + δ 1 θ − ( p − P ) 3 2 , that is, μ 2 δ − μ + δ 1 θ − ( p − P ) 2 > 2 δ + δ 1 θ − ( p − P ) 3 2 . It is easy to show that the graph of the left-hand side of this inequality is given in Figure 26. Then, the minimum tax rate increases the small country’s tax revenue if and only if 2 δ + δ 1 θ − ( p − P ) 3 < μ < 2 δ + δ 1 θ − ( p − P ) 2 + 2 δ + δ 1 θ − ( p − P ) 2 − 2 δ + δ 1 θ − ( p − P ) 3 , that is, 2 δ + δ 1 θ − ( p − P ) 3 < μ < 2 [ 2 δ + δ 1 θ − ( p − P ) ] 3 .

Figure 26:

Graph of the inequality in the proof of Case 5 of proof of Proposition 5 Part (ii).

So, for Case 5, we have μ ̄ = 2 [ 2 δ + δ 1 θ − ( p − P ) ] 3 .

Case 6: Assume that δ 1 θ < p − P < δ 1 θ + 2 δ . Then, we have t N = 2 δ + δ 1 θ − ( p − P ) 3 < T N = 2 δ 1 θ + δ + p − P 3 by Proposition 1 Part (iv), and cross-shopping in the large country by Proposition 2. These imply that we have R ( t N , T N ) = h 1 δ 2 δ 1 θ + δ + p − P 3 2 and r ( t N , T N ) = h 1 δ 2 δ + δ 1 θ − ( p − P ) 3 2 and the equilibrium is as given in Figure 27.

Figure 27:

Equilibrium in Case 6 of the proof of Proposition 5 Part (ii).

Following the discussion in Section 4, under a minimum-tax rate μ ∈ [t _N, T _N], we can ignore T(t) for t < μ. And t(T) will be equal to μ before μ-line cuts t(T) in its middle part, get longer before the discontinuity point, and will be equal to μ before the value of T is such that t(T) = μ. Then we have t m = μ < δ 1 θ + μ + ( p − P ) 2 = T m . There is a cross-border sale in the large country if and only if T _m P < t _n + p, that is, δ 1 θ − ( p − P ) < μ which always holds as δ 1 θ − ( p − P ) < 0 < t N < μ .

We have R ( μ ) = T m H + T m h t m + p − ( T m + P ) δ = h 1 δ δ 1 θ + μ + ( p − P ) 2 2 . Then, R(μ) ≥ R(t _N, T _N) if and only if h 1 δ δ 1 θ + μ + ( p − P ) 2 2 ≥ h 1 δ 2 δ 1 θ + δ + p − P 3 2 , that is, μ ≥ δ 1 θ + 2 δ − ( p − P ) 3 = t N which always holds as μ > t _N.

We have r ( μ ) = t m h 1 − t m + p − ( T m + P ) δ = h 1 δ μ 2 δ − μ + δ 1 θ − ( p − P ) 2 . Then, r(μ) ≥ r(t _N, T _N) if and only if h 1 δ μ 2 δ − μ + δ 1 θ − ( p − P ) 2 > h 1 δ 2 δ + δ 1 θ − ( p − P ) 3 2 , that is, μ 2 δ − μ + δ 1 θ − ( p − P ) 2 > 2 δ + δ 1 θ − ( p − P ) 3 2 . It is easy to show that the graph of the left-hand side of the above inequality is as given in Figure 28. Then, the minimum tax rate increases the small country’s tax revenue if and only if 2 δ + δ 1 θ − ( p − P ) 3 < μ < 2 δ + δ 1 θ − ( p − P ) 2 + 2 δ + δ 1 θ − ( p − P ) 2 − 2 δ + δ 1 θ − ( p − P ) 3 , that is, 2 δ + δ 1 θ − ( p − P ) 3 < μ < 2 [ 2 δ + δ 1 θ − ( p − P ) ] 3 .

Figure 28:

Graph of the inequality in Case 6 of proof of Proposition 5 Part (ii).

So, for Case 6, we have μ ̄ = 2 [ 2 δ + δ 1 θ − ( p − P ) ] 3 . □

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Received: 2022-09-01

Accepted: 2024-12-30

Published Online: 2025-01-13

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Articles in the same Issue

https://doi.org/10.1515/bejeap-2022-0332

Keywords for this article

commodity-tax competition; cross-border shopping; asymmetric prices; tax harmonization; minimum tax rate